Concept

Propagation — where it appears

The step of a phrase index's search that produces the occurrences lying inside a phrase from the ones they were copied from. Its cost is the phrases examined per known occurrence, and its bookkeeping is what a text-order traversal makes unnecessary.

Named by 5 essays across 5 fields — each of them below, with the objects they name alongside it.

occurrences reportedin text order16from the right2 — 14 lostevery one it reports is real, so the answer is short rather than wrong1 of them were found by the boundary search and never propagated16 copies · 6-character pattern87.5% lost

From the right, two of sixteen

Run the same sweep in the opposite direction and every occurrence it produces lands behind its own cursor. It reports two of sixteen, every one of them genuinely there, and nothing about the answer says fourteen are missing.

floors · Parse
01,024120 phrases · 1,024 characters60 drawn, every one pointing right

Every copy points right

A greedy self-referential parse chooses each phrase's source from text already produced, so an occurrence copied from another lies strictly to its right. That one fact removes a visited set from a propagation, exactly as a left-first walk removed an array one strand ago.

structures · Parse
"t than"8 found · set 8 · frontier 4"of her"7 found · set 7 · frontier 4" that "16 found · set 16 · frontier 11" every"31 found · set 31 · frontier 19"the ev"8 found · set 8 · frontier 4"o of c"8 found · set 8 · frontier 5" than "15 found · set 15 · frontier 9"ime ra"7 found · set 7 · frontier 4the visited set, pale; the sweep's frontier, dark17,715 phrases examined either way1.67x on what is held

The same occurrences, less bookkeeping

Two traversals examine identically many phrases and report identically many occurrences. What differs is that one holds every occurrence found so far in a set and the other holds an ordered list and a cursor.

indexes · Parse
0100200300400102030copies of the basebits held during one querythe visited set: 420the frontier: 2666-character patterns20x to 1.58x

A constant factor, not a term

The substitution was expected to remove a quantity proportional to the answer. Both traversals hold a quantity proportional to the answer, and what it removes is a factor of 1.6 that shrinks as the collection grows.

wrong · Parse
phrases examined per probethe linear scan207.0sorted, with a running maximum10.9 — 19xthe running maximum costs 2,821 bits — 21.6% of the index40 of 40 positions checked, and the two agree at all of them238 phrases · 6,144 characters19x on the scan

The scan the order does not touch

Two hundred and seven phrases examined per probe, against eleven. A running maximum of the source regions' right ends lets a leftward walk stop for good, and it costs twenty-two per cent more bits.

practice · Parse

Named alongside it

The objects these essays reach for when they reach for this one.

Phrase indexSecondary occurrenceLempel ziv parseTraversal orderVisited setIndex sizeWorking memoryCorrectnessInterval stabbingPrevious occurrence chainRejection testRunning maximum

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